Ô tô chuyển động trên đường cong có bán kính R = 300(m). Nếu vận tốc ôtô tăng đều từ v1 =

The distance between slits in a double-slit experiment is decreased by a factor of 2. If the distance between fringes is small c

Elodia [21]

Answer:

the distance between adjacent fringes is increased by a factor o 2

Explanation:

To find how the distance between fringes is modified you can use the following formula for the calculation of the distance between fringes:

$\Delta y=\frac{\lambda D}{d}$

D: distance to the screen

d: distance between slits

λ: wavelength of the light

if d is decreased by a factor of 2, that is d'=1/2d, you have:

$\Delta y'=\frac{\lambda D}{d'}=\frac{\lambda D}{(1/2)d}=2\Delta y$

hence, the distance between adjacent fringes is increased by a factor o 2

4 0

3 years ago

A 0.40 kg mass hangs on a spring with a spring constant of 12 N/m. The system oscillated with a constant amplitude of 12 cm. Wha

Vaselesa [24]

Answer:

The maximum acceleration of the system is 359.970 centimeters per square second.

Explanation:

The motion of the mass-spring system is represented by the following formula:

$x(t) = A\cdot \cos (\omega \cdot t + \phi)$

Where:

$x(t)$ - Position of the mass with respect to the equilibrium position, measured in centimeters.

$A$ - Amplitude of the mass-spring system, measured in centimeters.

$\omega$ - Angular frequency, measured in radians per second.

$t$ - Time, measured in seconds.

$\phi$ - Phase, measured in radians.

The acceleration experimented by the mass is obtained by deriving the position equation twice:

$a (t) = -\omega^{2}\cdot A \cdot \cos (\omega\cdot t + \phi)$

Where the maximum acceleration of the system is represented by $\omega^{2}\cdot A$ .

The natural frequency of the mass-spring system is:

$\omega = \sqrt{\frac{k}{m} }$

Where:

$k$ - Spring constant, measured in newtons per meter.

$m$ - Mass, measured in kilograms.

If $k = 12\,\frac{N}{m}$ and $m = 0.40\,kg$ , the natural frequency is:

$\omega = \sqrt{\frac{12\,\frac{N}{m} }{0.40\,kg} }$

$\omega \approx 5.477\,\frac{rad}{s}$

Lastly, the maximum acceleration of the system is:

$a_{max} = \left(5.477\,\frac{rad}{s})^{2}\cdot (12\,cm)$

$a_{max} = 359.970\,\frac{cm}{s^{2}}$

The maximum acceleration of the system is 359.970 centimeters per square second.

7 0

3 years ago

What are plants and animals that get their energy from eating other things?

Zanzabum

Answer:

B. Consumers.

Explanation:

4 0

3 years ago

Read 2 more answers

Read through the scenarios below and calculate the predicted change in kinetic energy of the object compared to a 100 kg ball tr

sashaice [31]

Okay, first off, the formula for Kinetic Energy is:

KE = 1/2(m)(v)^2

m = mass

v = velcoity (m/s)

Using this formula, we can then calculate the kinetic energy in each scenario:

1) KE = 1/2(100)(5)^2 = 1,250 J

2) KE = 1/2(1000)(5)^2 = 12,500 J

3) KE = 1/2(10)(5)^2 = 125 J

4) KE = 1/2(100)(5)^2 = 1,250 J

4 0

3 years ago

Read 2 more answers

Which of the following statements is true of chords? Select all that apply.

tankabanditka [31]

All of those are true, except the one about the radius.

Both ends of a chord have to be on the circle, but one end
of a radius is at the center, so a radius can't be a chord.

5 0

3 years ago